example of transcendental number
The following is a classical application of Liouville’s approximation theorem. For completeness, we state Liouville’s result here:
Theorem 1.
For any algebraic number α with degree m>1, there exists a constant c=c(α)>0 such that:
|α-pq|>cqm |
for all rationals p/q (with q>0).
Next we use the theorem to construct a transcendental number.
Corollary 1.
Proof.
Clearly, the number ψ is well defined, i.e. the series converges. Indeed,
110n!<110n |
and ∑∞n=110-n=1/9. Thus, by the comparison test, the series converges and 0<ψ<1/9.
Suppose, for a contradiction, that ψ is algebraic of degree m. We will construct infinitely many rationals p/q such that
|ψ-pq|<cqm |
where c=c(ψ) is the constant given by the theorem above. Let k∈ℕ be such that 1/2k<c. Then, in fact, we will show that there are infinitely many rationals p/q with q≥2 such that
|ψ-pq|<1qm+k<12k⋅1qm<cqm |
For all j>k+m we define a rational number pj/qj by:
pj=10j!j∑n=110-n!,qj=10j! |
then pj and qj are relatively prime integers and we have:
|ψ-pjqj| | = | ∞∑n=j+1110n! | ||
< | 110(j+1)!(1+110+1102+…) | |||
= | 10/9⋅1q(j+1)j | |||
< | 1qjj | |||
< | 1q(k+m)j |
where in the last inequality we have used the fact that j>k+m. Therefore, all rationals {pj/qj}∞j=k+m+1 satisfy the desired inequality, which leads to the contradiction with the theorem above. Thus ψ cannot be algebraic and it must be transcendental.
∎
Many other similar transcendental numbers can be constructed in this fashion.
Title | example of transcendental number |
---|---|
Canonical name | ExampleOfTranscendentalNumber |
Date of creation | 2013-03-22 15:02:45 |
Last modified on | 2013-03-22 15:02:45 |
Owner | alozano (2414) |
Last modified by | alozano (2414) |
Numerical id | 7 |
Author | alozano (2414) |
Entry type | Example |
Classification | msc 11J82 |
Classification | msc 11J81 |
Related topic | LiouvillesTheorem |
Related topic | RothsTheorem |